Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $t \neq 0$. $a = \dfrac{t + 7}{4t + 20} \times \dfrac{t^2 + 7t + 10}{8t^2 + 56t} $
Solution: First factor the quadratic. $a = \dfrac{t + 7}{4t + 20} \times \dfrac{(t + 5)(t + 2)}{8t^2 + 56t} $ Then factor out any other terms. $a = \dfrac{t + 7}{4(t + 5)} \times \dfrac{(t + 5)(t + 2)}{8t(t + 7)} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac{ (t + 7) \times (t + 5)(t + 2) } { 4(t + 5) \times 8t(t + 7) } $ $a = \dfrac{ (t + 7)(t + 5)(t + 2)}{ 32t(t + 5)(t + 7)} $ Notice that $(t + 7)$ and $(t + 5)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac{ \cancel{(t + 7)}(t + 5)(t + 2)}{ 32t\cancel{(t + 5)}(t + 7)} $ We are dividing by $t + 5$ , so $t + 5 \neq 0$ Therefore, $t \neq -5$ $a = \dfrac{ \cancel{(t + 7)}\cancel{(t + 5)}(t + 2)}{ 32t\cancel{(t + 5)}\cancel{(t + 7)}} $ We are dividing by $t + 7$ , so $t + 7 \neq 0$ Therefore, $t \neq -7$ $a = \dfrac{t + 2}{32t} ; \space t \neq -5 ; \space t \neq -7 $